I was given this function:
$$f(x)=\begin{cases}\displaystyle|x|^p\cos\Big(\frac\pi{|x|^q}\Big),&x\ne0\\0,&x=0\end{cases}$$
And was asked to find for what $p, q>0$ it is differentiable at $x=0$.
First I saw it is continuous when $p>0, q>0$.
Now, I tried to see if the limit for $f'(x)$ exists at $x=0$. This function is even so I looked at the right side only.
$$\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0^+}\frac{x^p\cos\Big(\displaystyle\frac\pi{x^q}\Big)}x=\lim_{x\to0^+}x^{p-1}\cos\Big(\frac\pi{x^q}\Big)$$
I get that this limit exists when $p>1$, and for all $q>0$, but looking at the graph online it doesn't seem to be right. What am I doing wrong here (if anything)?
Thanks a lot!
Your conclusion is right for $p>1$ and all values of $q>0$. The reason why you don't observe so on the graph is the the oscillation of the function increases around $x=0$ so it's indistinguishable to see whether the function is differentiable in $x=0$ or not. Also the function has no continuous derivative in $x=0$ for $0<p\le 1$. The figure below shows why: